# Prove the following equation

• Oct 25th 2012, 10:22 AM
Ogniok
Prove the following equation
Hi. How to prove that equation:

\$\displaystyle |AB|^{2} = |QA| * |AE| + |NB| * |BD|\$

\$\displaystyle |BN| = |BC|\$
\$\displaystyle |AQ| = |AC|\$

\$\displaystyle |AD|\$ and \$\displaystyle |BD|\$ are heights of acute-angled triangle \$\displaystyle ABC\$.

Illustration: http://i.imgur.com/9O8Bm.jpg

I would really appreciate any help. http://www.mymathforum.com/images/sm...on_biggrin.gif
• Oct 25th 2012, 02:05 PM
BobP
Re: Prove the following equation
First point is that most of the diagram is unnecessary.
If \$\displaystyle AQ=AC\$ you don't need \$\displaystyle Q\$ or \$\displaystyle P.\$
If \$\displaystyle BN=BC\$ you don't need \$\displaystyle N\$ or \$\displaystyle M.\$
Further, \$\displaystyle L\$ and \$\displaystyle K\$ serve no purpose.
All that you need is the triangle \$\displaystyle ABC\$ and the perpendiculars \$\displaystyle AD\$ and \$\displaystyle BE.\$

Now repeat that sequence using the triangles \$\displaystyle ABE\$ and \$\displaystyle BEC.\$

Having done that, add the resulting equation to the one that's just been derived.
• Oct 26th 2012, 06:29 AM
Ogniok
Re: Prove the following equation
Thank you very much. :D I've finally done that :D

Here are full calculations:

\$\displaystyle AB^{2} = AD^{2} + BD^{2} = (AC^{2} - CD^{2}) + BD^{2} = AC^{2} - (BC - BD)^{2} + BD^{2} = AC^{2} - (BC^{2} + BD^{2} - 2 * BC * BD) + BD^{2} = AC^{2} - BC^{2} + 2 * BC * BD\$

\$\displaystyle AB^{2} = BE^{2} + AE^{2} = (BC^{2} - CE^{2}) + AE^{2} = BC^{2} - (AC - AE)^{2} + AE^{2} = BC^{2} - (AC^{2} + AE^{2} - 2 * AC * AE) + AE^{2} = BC^{2} - AC^{2} + 2 * AC * AE\$

\$\displaystyle 2AB^{2} = AC^{2} - BC^{2} + 2 * BC * BD + BC^{2} - AC^{2} + 2 * AC * AE\$
\$\displaystyle 2AB^{2} = 2 * BC * BD + 2 * AC * AE\$
\$\displaystyle AB^{2} = BC * BD + AC * AE\$
• Nov 1st 2012, 02:02 AM
elisaevedent
Re: Prove the following equation